Reports of the Academy of Sciences of the USSR

1. Volume 145, No. 2

MATHEMATICS

A. ARKHANGELSKII

ON MAPPINGS OF METRIC SPACES

(Presented by Academician P. S. Aleksandrov on 26 II 1962)

Theorem 1. In order that a \(T_1\)-space \(Y\) have a uniform (see \((^1)\)) base, it is necessary and sufficient that \(Y\) be an open, continuous, and compact* image of some metric space \(X\); moreover, the space \(X\) can always be assumed to be zero-dimensional (in all senses) and to have the same weight as the space \(Y\).

This theorem is adjacent to known results of V. I. Ponomarev \((^2)\).

Lemma 1. Under an open bicompact mapping \(f\) of a space \(X\) onto \(Y\), an arbitrary open locally finite cover \(\eta\) of the space \(X\) is transformed into a point-finite cover \(f\eta\) of the space \(Y\).

Let us outline the proof of Theorem 1.

Necessity. With the aid of Lemma 1 we can, starting from some refining sequence of locally finite covers of the space \(X\), obtain a refining sequence of point-finite covers in the space \(Y\), and from this already follows (see \((^1)\)) the existence of a uniform base in the space \(Y\).

Sufficiency. In a space \(Y\) possessing a uniform base there exists a refining sequence of point-finite covers (see \((^1)\)): \(\varphi=\{\gamma_n\}\), \(\gamma_n=\{G_\alpha\mid \alpha\in M_n\}\).

Endow the sets \(M_n\) with the discrete topology and consider the Tikhonov product

\[ M=\prod_{n=1}^{\infty} M_n . \]

To a point \(m_0\in M\), \(m_0=\{\alpha_n^0\}\), we assign the point \(y_0\in Y\): \(f(m_0)=y_0\), if

\[ \bigcap_{n=1}^{\infty}G_{\alpha_n^0}\ni y_0, \]

and assign nothing if

\[ \bigcap_{n=1}^{\infty}G_{\alpha_n^0}=\Lambda . \]

Denote by \(E\) the set of points of the space \(M\) for which the mapping \(f\) is defined. Then one can verify that \(f\) is a single-valued, continuous, open, compact mapping of the zero-dimensional space \(E\) onto \(Y\).

From Theorem 1, with the aid of a theorem of P. S. Aleksandrov \((^1)\), it is derived that

Theorem 2. A collectively normal space \(Y\) that is a continuous open and compact image of a metrizable space is itself metrizable.

Theorem 3. Let \(X\) be an arbitrary metric space of weight \(\tau\). Then in \(B_\tau\)** there are subspaces \(M_1\) and \(M_2\) such that:
1) \(M_1\subseteq M_2\);
2) there exists a mapping \(f\): \(fM_2=X\), \(fM_1=X\), continuous on \(M_2\) (and hence on \(M_1\)) and open if it is considered only on \(M_1\), and closed on \(M_2\);
3) the mapping \(f\) is compact both on \(M_1\) and on \(M_2\);
4) if it is assumed that \(X\) is a complete metric space, then \(M_2\) can be chosen closed in \(B_\tau\).

Finally, we note that with the aid of Lemma 1 one easily proves

Theorem 4. An open continuous bicompact image of a paracompact space is a weakly paracompact space.

Remark 1. With regard to Theorems 2 and 3 one should point to interesting results of V. Ponomarev \((^2)\) and E. Michael \((^3)\).

* A mapping \(f:X\to Y\) is called compact (bicompact) if, for an arbitrary point \(y\in Y\), the set \(f^{-1}(y)\) is compact (bicompact).

** The space \(B_\tau\) is the space of the Tikhonov product of a countable set of discrete sets of cardinality \(\tau\).

Remark 2. P. S. Aleksandrov’s question: “Does there exist a nonmetrizable normal space possessing a uniform base?” can, after Theorem 1, be formulated as follows: “Will a normal space that is an open, continuous, and compact image of some metrizable space always be metrizable?”

A partial answer to this question is given by

Theorem 5. If a regular locally finally compact space \(Y\) is the image of some metrizable space \(X\) under an open, continuous, and compact mapping, then \(Y\) itself is metrizable.

Remark 3. In the statement of Theorem 2, compact mappings cannot be replaced by \(S\)-mappings (see \((^2)\)). However, the following is true.

Theorem 6. If a bicompactum \(\Phi\) is an open, continuous \(S\)-image of some metrizable space \(X\), then both \(\Phi\) and \(X\) have a countable base.

We introduce the notion of a regular mapping of a metric space. (This notion arose in connection with the notion of a weakly uniform mapping of a metric space, defined by V. Ponomarev in \((^2)\).)

Definition 1. A mapping \(f\) of a metric space \(X\) onto \(Y\) is called regular if for every point \(y \in Y\) and its neighborhood \(O_y \ni y\) there is a neighborhood \(O_1y \ni y\) such that
\[ \rho[f^{-1}(O_1y), X \setminus f^{-1}(O_y)] > 0. \]

Definition 2. A mapping \(f\) of a topological metrizable space (without a fixed metric) is called regular if there exists on \(X\) a metric with respect to which \(f\) is regular (in the sense of Definition 1).

There exist examples: 1) of a regular continuous one-to-one mapping of a metric space onto an arbitrarily bad space; 2) of a continuous two-to-one and, hence, compact mapping of a metric space \(X\) (a subset of the line) onto a metric space, which is not regular with respect to the metric prescribed on \(X\).

Theorem 7. A continuous mapping of a metrizable space \(X\) onto a metrizable space \(Y\) is always regular.

In the proof of this theorem an essential role is played by

Lemma 2. Let \(\varphi=\{\gamma_n\}\) be a refining sequence of locally finite covers of a normal\(^*\) space \(X\). Then there exists on \(X\) a metric \(\rho\) such that if \(\rho(x,y)<1/n\), where \(n\) is an arbitrary natural number and \(x,y\) are arbitrary points of the space \(X\), then there is an element \(U\in\gamma_n\) for which \(x\in U\) and \(y\in U\). (We shall call this metric “the metric associated with the system \(\varphi=\{\gamma_n\}\)”.)

Theorem 8. Let \(f\) be a continuous closed mapping of a metrizable space \(X\) onto a \(T_1\)-space \(Y\). Then, in order that the space \(Y\) be metrizable, it is necessary and sufficient that the mapping \(f\) be regular.

Theorem 9. Let \(f\) be a continuous open mapping of a metrizable space \(X\) onto a \(T_1\)-space \(Y\). Then, in order that the space \(Y\) be metrizable, it is necessary and sufficient that \(f\) be regular.

From Theorem 8 one can derive the following

Corollary 1. Let a continuous decomposition \(Z\) of a metrizable space \(X\) into closed sets with bicompact boundary be given. Then there exists on \(X\) a metric such that for every element \(\Phi \in Z\) and every closed subset \(F\) lying inside \(\Phi\), \(F \subseteq \langle\Phi\rangle\), one has
\[ \rho(F, X \setminus \Phi)>0. \]

Proof of Theorem 9. Necessity. Let \(\varphi=\{\gamma_n\}\) be a fundamental family of covers of the space \(X\), defined as follows:
\[ \gamma_n=\{G\mid G \text{ is open in } X,\ \operatorname{diam}G<1/n\}. \]

\(^*\) Instead of normality of \(X\) in the hypothesis of Lemma 2, one could, of course, assume only that \(X\) is regular.

We shall show that then \(\psi=\{\lambda_n=f\gamma_n\}\) is a fundamental set of coverings of the space \(Y\).

Let a point \(y\) and its neighborhood \(Oy\) be given arbitrarily. Since the mapping \(f\) is regular, there is a neighborhood \(O_1y\) of the point \(y\) for which
\(\rho[f^{-1}(O_1y),\, X\setminus f^{-1}(Oy)]>0\). Choose \(N\) so that
\[ \frac{1}{N}<\rho[f^{-1}(O_1y),\, X\setminus f^{-1}(Oy)]. \]
Now, if \(U\in\gamma_n\) and \(U\cap f^{-1}(O_1y)\ne\Lambda\), then \(U\subseteq f^{-1}(Oy)\). We shall show that \(\lambda_N(O_1y)\subseteq Oy\). Indeed, let \(V\in\lambda_n\) and \(V\cap O_1y\ne\Lambda\). Then \(U=f^{-1}(V)\in\gamma_n\) and, obviously,
\[ U\cap f^{-1}(O_1y)=f^{-1}(V)\cap f^{-1}(O_1y)=f^{-1}(V\cap O_1y)\ne\Lambda. \]
Then, by what has been proved, \(f^{-1}(V)\subseteq f^{-1}(Oy)\). Consequently, \(V\subseteq Oy\), or, in other words, \(\lambda_N(O_1y)\subseteq Oy\). This proves that \(\{\lambda_n\}\) is a fundamental set of coverings of the space \(Y\). By the theorem proved in \((^4)\), the space \(Y\) is metrizable. The necessity of the condition of the theorem is proved. The sufficiency is obvious.

Lemma 3. If the spaces \(X\) and \(Y\) are metrizable, and the mapping \(f:X\) onto \(Y\) is continuous and open, then there exists in \(X\) a complete sequence \(\{\gamma_n\}\) of locally finite coverings of the space \(X\), such that \(\{\lambda_n\}=\{f\gamma_n\}\) is a fundamental set of coverings of the space \(Y\).

Let us turn to the proof of the sufficiency of the theorem. Let the conditions of the theorem be satisfied, and let the systems \(\{\gamma_n\}\) and \(\{\lambda_n\}\) be chosen in accordance with the lemma. Introduce in \(X\) the metric \(\rho\) associated with the complete system of locally finite coverings \(\{\gamma_n\}\) (see Lemma 2). We shall show that the mapping \(f\) is regular with respect to this metric.

Indeed, let an arbitrary point \(y_0\in Y\) and its neighborhood \(Oy_0\ni y_0\) be taken. Consider the sets \(O\Phi_0=f^{-1}(Oy_0)\) and \(\Phi_0=f^{-1}(y_0)\). There is a neighborhood \(O_1y_0\) for which, for some natural number \(n_0\),
\[ \lambda_{n_0}(O_1y_0)\subseteq Oy_0. \]
Then for this \(n_0\), for any \(U\in\gamma_{n_0}\), from \(U\cap f^{-1}(O_1y_0)\ne\Lambda\) it follows that \(U\subseteq O\Phi_0\). Indeed, if \(U\cap f^{-1}(O_1y_0)\ne\Lambda\), then \(f(U)\cap O_1y_0\ne\Lambda\). But \(f(U)\in\lambda_{n_0}\), and therefore \(f(U)\subseteq Oy_0\) by the choice of the neighborhood \(O_1y_0\). Hence \(U\subseteq f^{-1}(Oy_0)\), or
\[ U\cap\bigl(X\setminus f^{-1}(Oy_0)\bigr)=\Lambda. \]
We shall show that
\[ \rho\bigl(X\setminus f^{-1}(Oy_0),\, f^{-1}(O_1y_0)\bigr)\ge \frac{1}{n_0}, \]
which will complete the proof of the theorem. But this is so, for if \(x_1\in X\), \(x_2\in X\) are arbitrary points such that \(x_1\in f^{-1}(O_1y_0)\) and \(x_2\in X\setminus f^{-1}(Oy_0)\), then, by what has been proved, from \(U\in\gamma_{n_0}\) it follows that either \(U\cap x_1=\Lambda\), or \(U\cap x_2=\Lambda\). Then, by the special property of the metric \(\rho\) (see Lemma 2), \(\rho(x_1,x_2)\ge 1/n_0\). But in that case
\[ \rho\bigl(X\setminus f^{-1}(Oy_0),\, f^{-1}(O_1y_0)\bigr) = \inf_{\substack{x_1\in f^{-1}(O_1y_0)\\ x_2\in X\setminus f^{-1}(Oy_0)}} \rho(x_1,x_2) \ge \frac{1}{n_0}>0, \]
which means that the mapping \(f\) is regular (with respect to the metric \(\rho\)). The theorem is proved.

The author expresses deep gratitude to Acad. P. S. Aleksandrov for his attention to the work.

Note added in proof. Recently I obtained a result which essentially generalizes Theorems 8 and 9:

Let \(f:X\to Y\) be such a continuous regular mapping of a metric space \(X\) onto a \(T_1\)-space \(Y\) that \(\langle fU_\varepsilon(f^{-1}y)\rangle\ni y\), i.e., the interior of the image of every \(\varepsilon\)-neighborhood of the full preimage of an arbitrary point \(y\in Y\) contains \(y\). Then the space \(Y\) is metrizable.

Obviously, if the mapping \(f\) is open or closed, then it all the more satisfies the condition of our theorem.

Received
23 II 1962

REFERENCES

\(^{1}\) P. Aleksandrov, Bull. Polish Acad. Sci., Ser. Math., Astr. and Phys., 8, 135 (1960).
\(^{2}\) V. Ponomarev, ibid., 8, 127 (1960).
\(^{3}\) E. Michael, Duke Math. J., 26, No. 4, 47 (1959).
\(^{4}\) A. Arkhangel’skii, DAN, 141, No. 1, 13 (1961).